Relations Lecture 9 - PowerPoint PPT Presentation

Relations Lecture 9


 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Symmetric: 
 Anti-symmetric: 
 Transitive: 
 Reflexive: 
 Irreflexive: 
 Only self-loops & No bidirectional Path from a to b All self-loops No self-loops bidirectional edges edges implies edge (a,b) The complete relation R = S × S is reflexive, symmetric and transitive Reflexive closure of R: Smallest relation R’ ⊇ R s.t. R’ is reflexive 
 Symmetric closure of R: Smallest relation R’ ⊇ R s.t. R’ is symmetric 
 Transitive closure of R: Smallest relation R’ ⊇ R s.t. R’ is transitive Each of these is unique


 Question 1 Let ⊏ be the empty relation (i.e., ∀ a,b ¬(a ⊏ b)). 
 Choose the best option. 
 A. ⊏ is transitive 
 B. ⊏ is irreflexive 
 C. ⊏ is symmetric 
 All. Also, anti-symmetric. D. All of the above 
 E. Some of the above (but not all)


 Question 2 Let ⊏ be the relation over integers defined as 
 x ⊏ y if |x-y| ≤ 10. Choose the best option. 
 A. ⊏ is transitive 
 B. ⊏ is reflexive 
 C. ⊏ is symmetric 
 Not transitive D. All of the above 
 E. Some of the above (but not all)

Equivalence Relation A relation that is reflexive, symmetric and transitive e.g. is a relative, has the same last digit, is congruent mod 7, … Claim: Let Eq(x) ≜ {y|x ～ y}. If Eq(x) ∩ Eq(y) ≠ Ø, then Eq(x) = Eq(y). Let z ∈ Eq(x) ∩ Eq(y). ∀ w ∈ Eq(x), x ～ w. Also, x ～ z ⇒ w ～ z. 
 Also, y ～ z ⇒ y ～ w ⇒ w ∈ Eq(y). i.e., Eq(x) ⊆ Eq(y). The Equivalence classes partition the domain Square blocks along the P 1 ,..,P t ⊆ S 
 diagonal, after sorting s.t. 
 the elements by “Cliques” for P 1 ∪ .. ∪ P t = S 
 equivalence class each class P i ∩ P j = Ø


 
 
 Question 3 Which one(s) represent(s) equivalence relation(s) 
 R 1 R 2 R 3 
 
 not transitive not reflexive A. R 1 and R 3 
 B. R 1 only 
 C. R 2 only 
 D. R 3 only 
 E. None of the above

Strict partial order: Posets irreflexive, rather than reflexive Partial order: a transitive, anti-symmetric and reflexive relation e.g. ≤ for integers, divides for integers, ⊆ for sets, “containment” for line-segments Partial: Some pair may be “incomparable” Cyclic: Some node s.t. Transitive and anti-symmetric → “acyclic” you can leave it through an edge (not self-loop), Partially ordered set (a.k.a Poset): 
 move through some a set and a partial order over it edges, and return to the node S 1 ={0,1,2,3}, S 2 ={1,2,3,4}, Check: 
 - Anti-symmetric (no bidirectional 
 S 3 ={1,2}, S 4 ={3,4}, 
 edges), 
 S 5 = {2}. 
 - Transitive, 
 Relation ⊆ - Reflexive (all self-loops)

Posets Do exist in finite posets 
 (Prove by induction on |S|) Maximal & minimal elements of a poset (S, ≼ ) x ∈ S is maximal if ∄ y ∈ S-{x} s.t. x ≼ y x ∈ S is minimal if ∄ y ∈ S-{x} s.t. y ≼ x Need not exist (e.g., in ( Z , ≤ )). Need not be unique when it exists (e.g., (S, ⊆ ), where S is the set of all subsets of integers that have at least one odd number) Greatest element in T ⊆ S: x ∈ T s.t. ∀ y ∈ T, y ≼ x 
 Least element in T ⊆ S: x ∈ T s.t. ∀ y ∈ T, x ≼ y Need not exist, even if T finite. Unique when it exists. Upper Bound for T ⊆ S: x s.t. ∀ y ∈ T, y ≼ x 
 Lower Bound for T ⊆ S: x s.t. ∀ y ∈ T, x ≼ y Least Upper Bound for T: Least in {x| x u.b. for T} 
 Greatest Lower Bound for T: Greatest in {x| x l.b. for T}

An Example Let a ⊏ b iff b/a is prime (with Z + as the domain) Let ≼ be the transitive and reflexive closure of ⊏ a ≼ b iff a|b 16 8 12 Divisibility poset: ( Z + , ≼ ) 4 6 9 10 14 15 When is c a lower bound 
 2 3 5 7 11 13 for T={a,b}? c ≼ a and c ≼ b. 1 c is a common divisor for {a,b}. gcd(a,b) = greatest lower bound for {a,b} in this posey


 
 
 Total/Linear Order In some posets every two elements are “comparable”: for {a,b}, either a ⊑ b or b ⊑ a Can arrange all the elements in a line, with all possible right-pointing edges (plus, self-loops) 
 If finite, has unique maximal and unique minimal elements (left and right ends)

Order Extension A poset P’=(S, ≤ ) is an extension of a poset P=(S, ≼ ) if ∀ a,b ∈ S, a ≼ b → a ≤ b Any finite poset can be extended to a total ordering (this is called topological sorting) By induction on |S| Induction step: Remove a minimal element, extend to a total ordering, reintroduce the removed element as the minimum in the total ordering. For infinite posets? The “Order Extension Principle” is typically taken as an axiom! (Unless an even stronger axiom called the “Axiom of Choice” is used)

Chains Finite if S is finite C ⊆ S is called a chain if ∀ a,b ∈ C, either a ≼ b or b ≼ a That is, (C, ≼ ) is a total order 16 8 12 Every element a ∈ S belongs 
 4 6 9 10 14 15 to some chain in which it is 
 2 3 5 7 11 13 the maximum element 
 1 (possibly just {a}) Height(a) = max length chain with a as the maximum E.g., In “Divisibility poset,” height(1)=1, height(p)=2 for all primes p. For m=p 1d1 ⋅ … ⋅ p tdt (p i primes) height(m) = 1+ Σ i d i

Anti-Chains A ⊆ S is called an anti-chain if 
 ∀ a,b ∈ A, a ≠ b → neither a ≼ b nor b ≼ a 16 (A, ≼ ) is the equality relation 8 12 Let A h = { a | height(a)=h } 4 6 9 10 14 15 2 3 5 7 11 13 For every finite h, A h is an 
 1 anti-chain (possibly empty) Otherwise, ∃ a ≠ b, a ≼ b with height(a) = height(b) = h. 
 height(a) = h ⇒ ∃ chain C s.t. a=max(C) and |C|=h 
 ⇒ b ∉ C and C’=C ∪ {b} is a chain with b=max(C’) 
 How? ⇒ height(b) ≥ h+1 !

Anti-Chains A ⊆ S is called an anti-chain if 
 ∀ a,b ∈ A, a ≠ b → neither a ≼ b nor b ≼ a 16 (A, ≼ ) is the equality relation 8 12 Let A h = { a | height(a)=h } 4 6 9 10 14 15 2 3 5 7 11 13 For every finite h, A h is an 
 1 anti-chain (possibly empty) In a finite poset, since every element has a finite height, every element appears in some A h : i.e., A h s partition S Mirsky’ s Theorem: Least number of anti-chains needed to partition S is exactly the length of a longest chain Height of the poset